# What method can be used to solve this system differential equations with term cos(y(t))?

I have the following system \begin{gathered} \frac{{dx}}{{dt}} = A*cos(y) + B*cos (b*t) \hfill \\ \frac{{dz}}{{dt}} = A*sin(y) - Ccos (c*t) \hfill \\ \frac{{dy}}{{dt}} = D*tan(d*t), \hfill \\ \end{gathered}

where $$A,B,C,D,b,c,d$$ are constants.

$$t \in (0,T)$$, where $$T < \infty$$

I do not have any idea what approach to use to solve it

What I have done is as follows: 1) Integration of y $$y = D{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|_{t\max }} - D{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|_{t\min }}$$ 2)I inserted y into the equation with x $$x = A\int {\cos \left( {D{{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|}_{t\max }} - D{{\left. {\frac{{ln(1 + t{an^2}(d*t))}}{{2d}}} \right|}_{t\min }}} \right)} dt + smth\_simple$$

But this is very fastly oscillating function and limits of integral confuse me. Notice that $$\left| {t\max - t\min } \right| \leqslant 10\sec$$ P.S. y can change only within 10 sec period, then it is constant

Probably I should assign t to tmax . And if tmin is equal to 0. I can integrate x from 0 to something. But what if tmin is not zero?

• I'll edit my post to show Aug 1 '17 at 18:42
• And what is $g$? Aug 1 '17 at 18:52
• tg a.k.a. tan a.k.a. Tangent function Aug 1 '17 at 18:59
• You mean the following? $$\dfrac{dy}{dt} = D \tan (d \times t)$$
– Moo
Aug 1 '17 at 19:12
• Yes, this is what I mean Aug 1 '17 at 19:15

If $g(t)$ is a known function, integrate the equation for $dy/dt$ to get $y$, then substitute the result into the equations for $dx/dt$ and $dz/dt$ and integrate.